Question

The velocity of a particle moving along x-axis is given as \( V = x^2 - 5x + 4 \) (in m/s) where x denotes the x-coordinate of the particle in metres. The magnitude of the acceleration of the particle when the velocity of the particle zero is

• A. 2 m/s\(^2\)
• B. 3 m/s\(^2\)
• C. Zero
• D. 1 m/s\(^2\)

Hint:

When velocity is given as a function of position, use: \[ a = v \frac{dv}{dx} \] If velocity becomes zero, acceleration also becomes zero (unless the expression is undefined).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The velocity of the particle is given as a function of position: \[ v = x^2 - 5x + 4 \] We are asked to find the magnitude of acceleration when the velocity becomes zero.
Step 2: Key Formula or Approach:
Acceleration can be written as: \[ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx} \] Step 3: Detailed Explanation:
First, find when velocity is zero: \[ x^2 - 5x + 4 = 0 \] \[ (x - 1)(x - 4) = 0 \Rightarrow x = 1 \text{ or } x = 4 \] Now, compute: \[ \frac{dv}{dx} = 2x - 5 \] Acceleration: \[ a = v \cdot (2x - 5) \] At both \( x = 1 \) and \( x = 4 \), we already know: \[ v = 0 \] So, \[ a = 0 \cdot (2x - 5) = 0 \] Step 4: Final Answer:
The magnitude of acceleration is: \[ \boxed{0} \]